Question: Let $f$ be a differentiable function with $f(7)=-2$ and $f'(7)=5$. What is the value of the approximation of $f(6.9)$ using the function's local linear approximation at $x=7$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2.7$ (Choice B) B $-2.6$ (Choice C) C $-2.5$ (Choice D) D $-2.4$
Solution: The local linear approximation of $f$ at $x=7$ is achieved using the equation of the line tangent to $f$ at $x=7$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=f'(7)(x-7)+f(7)$. Plugging $f(7)=-2$ and $f'(7)=5$, we obtain $L(x)=5(x-7)-2$. To approximate $f(6.9)$, all we need is to plug $x=6.9$ into $L(x)$. $\begin{aligned} L(6.9)&=5(6.9-7)-2 \\\\ &=5(-0.1)-2 \\\\ &=-2.5 \end{aligned}$ In conclusion, the approximation of $f(6.9)$ using the function's local linear approximation at $x=7$ is $-2.5$.